text:
Does sudden downdraft increase or reduce drag?
Let say wing section in wind tunnel operate at 10° AoA, somehow we force downdraft that reduce wing effective AoA to 6°.
Of course lift sensor will show reduction in lift.
Will drag sensor show increase  or reduce in drag?

In the steady state a 6 degree AoA Will normally have less drag than 10 degrees unless you have a very unusual section.  However, what will happen during the transition depends on the nature of the airflow at that point, which is uncertain.

I'm not sure what you're referring to.  If AoA is reduced until lift is zero then there should be no tip vortices and so no induced drag, only parasitic.

If the change in AoA is fast enough, then strange behaviours start to appear.
The usual aerodynamic characteristics like $C_l$ or $C_d$ which are seen in any standard plot are obtained modifying the AoA of the airfoil in a quite slow way. For example, for a conventional NACA 0012 they look something like that (plot from this answer):

This way of testing reveals only the so called "static" aerodynamic characteristics of the airfoil. Anyway, if the AoA is changed relatively faster, then those plots change quite drastically. Said $\omega$ the frequency at which the airfoil is made oscillate in the test section, a number called "reduced frequency" can be defined as:
$k=\frac{\omega c}{2V}$
Every time that $k$ is bigger than some 0.05, then also "unsteady" aerodynamic phenomena emerge which cannot be disregarded. Unsteady aerodynamic phenomena with $k$ bigger than 0.05 are normally encountered on rotary wings (helicopter and wind turbine rotors) airplane wings at high speed (aka flutter) and civil structures in strong wing (light poles, bridges, power lines, ...).
The following plot (from this NASA technical note plus a couple of arrows from my side) shows how the lift coefficient changes when the same NACA 0012 of before is made oscillate at $k=0.065$:

Now the following can be seen:

increasing the AoA, $C_l$ increases but  following a line (blue arrow) which is higher than the static line (in black); the dynamic lift coefficient is higher than the static one, the AoA being the same;
when AoA decreases, lift decrease but following yet another line (red arrow) which lies lower than the static one; the dynamic coefficient is lower than the static one, the AoA being the same;
the lift coefficient at stall is now more or less 2.5, up from the 1.5 for the static condition; that's a lot more!

This behaviour is termed "dynamic stall". Physically, what happens is very similar to how a delta wing generate a quite high lift at subsonic speeds: a vortex is continuously shed from the leading edge and travels down the upper surface till the trailing edge; this vortex helps the airflow in remaining attached to the surface even beyond (static) stall.

"Physically, what happens is very similar to how a delta wing generate a quite high lift" - no, not at all. What really happens is that the boundary layer at the rear of the airfoil comes from flow at low AoA while the pressure at the forward part comes from the suddenly high AoA. Separation is delayed and stall shifted upwards. The opposite happens on the way down: Now the thick boundary layer of the high AoA flow reduces lift at low AoA. Rinse and repeat.

@PeterKämpf: I agree that the phenomenon is more complicated than wrhat I have written, but the main source of the peak in the plot is the vortex shed at the leading edge. Otherwise Leishman is wrong and you should correct him

OK, some very interesting answers about transient effects and lift theory, but what happens in the next few seconds?

a downdraft can be considered an "anti-thermal"

A downdraft will cause an aircraft to lose altitude unless power is increased.  Here, the effect on fuel consumption is identical to increased drag.
Aerodynamicly, the shift in relative wind may cause a brief reduction in drag and lift.  However, the pilot must compensate by increasing AOA (to where it was originally), as well as power, to maintain altitude.
"Sudden downdraft" at its extreme is seen in microbursts, which are very dangerous at low altitudes above ground level.

Additionally, sudden downdraft may cause all kinds of turbulence before the new relative wind is established.  I would not bet on much in the way of drag reduction.

@ Robert DiGiovanni agreed - the OP is asking about a downdraft specifically rather than a change in the angle of the wing itself.  depending on the cause, the airflow may be relatively smooth or could be highly turbulent.

@Frog yea, the whole moving with the airmass thing works against the plane pretty fast.  Essentially, one must climb to hold altitude after recovering AoA.  Sophits graphs are interesting as to what happens exiting the downdraft.


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Can vane delay flow separation?
In general is a slat better  than a vortex generator to delay flow separation, for high AoA or extremely curvature wing/objects?
Does this "roof vane spoiler" works same as slat at wing, because the position of vane is so far back compared to the slat  and where it gets "new fresh air" like a real slat on a wing?
Can we use this vane so much back at some extremely high curve wing to delay stall?

I doubt this vane can help to stay flow attached.

In what case? Are you asking about a car or a wing? How are the pictures you show applicable to your question?

He wants to know if VGs on the roof of the car ahead of the window would do the same job as those appendages.

Of course they wouldn't. Vortex Generators don't look nearly as cool as a spoiler, so they wouldn't meet the requirement that the spoiler is fulfilling - looking impressive.

@RalphJ have you ever seen the row of VGs on the Mitsubishi Evo 7? They looked pretty dog-gone cool in a day when everyone else had "just" that little vane at the top of the rear window!

On airplanes, slats are definitely more effective. A slatted wing's stalling AOA will go from 15-16 degrees to mid 20s.
VGs on the leading edge, if placed at the right spot (just forward of the laminar separation point at high AOA), will increase the stall AOA about half that, to somewhere around 20 degrees.
Problem is, those things on the cars aren't really slats, which have to go at the leading edge.  Those are more properly called guide vanes, which are not quite the same thing, aerodynamically.
You'd want to see a wind tunnel smoke test to demonstrate that something is happening, and they are not just sitting in the already turbulent air doing nothing.  Or you could stick some tufts to the back window and observe them as you drive with and without the vanes.
You would never put guide vanes like that on a wing though, as the drag they produce would far outweigh any benefit, and they may produce nasty stall characteristics.

You think that vane dont work,just stlying?

My hunch is it's mostly styling, or someone thinks it works but if you tested it maybe not so much, and for it to actually work it would need to be much larger.   Tufts taped to the back window would show a pretty noticeable difference, so easy to test it out.

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text:
What is max angle of surface without stall?
AoA is by defintion angle between choord line and airflow, here is 25°.
But max effective AoA here is 60°, so flow will separate for sure here..
What is max lets call it "effective AoA" that air can follow, not include multi element wing, slats and  VG.
Use diffuser as example, but can apply on very high curvature wing/sail too.

Wouldn’t the answer be airfoil specific?

@MichaelHall Yes it would but maybe some general angle that airflow can handle without help from VGs, slats, multi wings..

The graph of lift vs AoA for a higher cambered wing is simply shifted to the left.  That wing will start generating lift at negative AoA.
The behavior of the flow on the ever important top rear of the wing will depend on fluid density, viscosity, and flow rate.
As an airfoil, this wing will be very draggy, most likely outside designs considered for the application.  Decrease camber and lengthen chord an you have a nice low speed lifting wing.  Fill in the bottom at higher airspeed (Reynolds number) and the wing can produce the same amount of lift at a higher speed with less drag (fuel) per mile.

what is the max angle of surface without stall?

This will depend on the local airflow, as well as the free stream.  The picture essentially shows an inverted wing with a flaps fully extended.  We would expect the 60 degree section to produce lots of drag.  Less deflection would reduce drag and allow that portion of the wing to continue producing lift efficiently, in spite of its increased angle to the free stream.
Eventually, turbulence caused by differences in free stream and local flow will start to show as pre-stall buffeting.  A designer can tailor the curve of the upper wing to avoid this as much as possible.
Notice that sails do not emphasize lift as much as thrust, and are commonly used at much higher Angles of Attack.

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text:
Is air less dense at the upper wing surface,for subsonic speeds?
If pressure drop at top wing side,density of air must also decrease because density depend on pressure and temperature.
Same apply on the bottom side where density must increase.
So why density effetcs caluculate only from Mach 0.3 above?

A flow can be considered incompressible if the variation in density is small. As you said, density is proportional to pressure and the inverse of temperature. The static pressure at sea level in the standard atmosphere is 101300 Pa. The dynamic pressure at Mach 0.3 (roughly 100 m/s) is
$p_{dyn} = \frac{\rho} {2} V^2 = \frac{1225\;\textrm{kg/m}^3}{2}(100\;\textrm{ m/s})^2 = 6125\;\textrm{Pa}$
So the highest total pressure you can get would be only 6 % higher than the ambient pressure, and the lowest one depends on the shape of the airfoil, but is typically of the same magnitude.
For most applications, this variation of ~5 % is small enough to be negligible.


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Does FAA have any info on army air corps pilots during wwII (know the FAA was formed later)?
Did the FAA (or its precursors) have any relationship to Army Air Corps pilots during WW2?  If so, how would I request information on my deceased father.  Most of his military records did not survive the fire in St. Louis.

Unfortunately no, the FAA does not certify or track military pilots. The only records the FAA has or carried over from CAA would have is if he converted his hours for a certificate, but I would be surprised it would be anything more than a number.

FAA does not.
Each military branch has many, many records, which are often well preserved but poorly sorted. They also have history offices, liaison offices, and many other offices besides which may have access to those records. There are also often amateur or professional historians who have information about a great many specific units.
You are more likely to get a good answer if you provide as much information as you know about your father's service here, and more likely to get a pointer to where to ask a government organization if you include even minimal information (What did he fly? Where? Did he mention any organization lower than the Army Air Corps? Who have you tried to contact? Did he show you any medals that might have public writeups? Etc)


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If I get my Private Pilots License before college, how far will that set me ahead, if at all?
So I am looking into getting my PPL and I was wondering if there was any benefit when it comes to college. I am wanting to go to Purdue, preferably, and major in Professional Flight. If I went ahead in getting my PPL now would that get me ahead in college? If it doesn't I don't think it would be worth spending all that money.

Well, since you want to go to Purdue specifically, a good place to start would be the Purdue Professional Flight website.  I found this there:

FAQ: Students Coming to School with a Private Pilot’s Certificate. Is there an advantage?

Depending on each individual situation, a student coming in with their Private Pilot’s Certificate can be either an advantage or disadvantage. The only real difference is cost. Sometimes it is more cost efficient for a student to get their certificate from Purdue or an outside flight school. It is ultimately up to the student.

Well that's one of the most non-answer answers I've seen! Good find.

@Greg Hewgill: Though the problem's not with the answer, but with real life.  If your family owns a plane, it's probably a lot cheaper to get your PPL in it, than if you have to rent one from an FBO.


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Calculating the horizontal drift for parachute, is it okay to assume the "starting speed" to be terminal velocity?
Plus in the worst case of the maximum range, can we assume we are already at the terminal velocity when we're deployed at 600m?
If we assume that we are deployed at 600m, we "freefall" for a few seconds while the parachute completely opens, catches air and reaches terminal velocity.
So then its just a case of balancing the total drag (from both components - falling and sidewind) with the estimated mass of the payload?

Unless dropping from a stationary delivery platform, your package will have a significant horizontal velocity, you must account for this, both in drift and chute opening time.

The payload is basically a bomb until the canopy is fully open, following a ballistic trajectory, except that the it's accelerating then decelerating during the deployment, so that has to be accounted for.  If the drop height is several thousand feet, it might be insignificant enough to ignore or just add a ball park fudge factor, but if the drop is very low and you need extreme precision, it can be a quarter of the total estimate.

Paratroopers who jump with a static line out of a C-17 or C-130 are essentially at terminal velocity as soon as they're out of the aircraft. Those who jump from a Blackhawk (or  B.A.S.E. jumper), not so -  they are accelerating toward terminal velocity at the point that their chute starts to deploy. Different cases, different assumptions.

Also, to be clear on terminology, "terminal velocity" of the jumper/object without parachute is far higher than the terminal velocity with parachute. Jumping from a fast aircraft, you may start out near/at/above both of these speeds; the BASE jumper probably never reaches the former, and slows to the latter in less time than his jet-jumping compadre.


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What does "the other's right" refer to in 14 CFR Part 91.113?
There's a part in 14 CFR Part 91.113 (regarding airborne right-of-way rules) that's been confusing me:

Converging. When aircraft of the same category are converging at approximately the same altitude (except head-on, or nearly so), the aircraft to the other's right has the right-of-way.

Specifically, I don't fully understand the phrase "the aircraft to the other's right" - it would seem both aircraft will either perceive themselves as being to the left or right of the other meaning either both aircraft or neither aircraft would have the right of way. This is obviously not what is intended in FAR, so could someone give a more intuitive explanation of this? Thanks!

"It would seem both aircraft will either perceive themselves as being to the left or right of the other..."  Yes.  One is perceived to be on the left, and one is perceived to be on the right.  And it is the one to the right that has the right of way.  (I guess I just don't see where the confusion lies... )

You wrote that:

it would seem both aircraft will either perceive themselves as being to the left or right of the other meaning either both aircraft or neither aircraft would have the right of way.

But that's only the case if the two aircraft are approaching head-on, or nearly so, and the FAR you quoted already accounts for that situation:

Converging. When aircraft of the same category are converging at approximately the same altitude (except head-on, or nearly so), the aircraft to the other's right has the right-of-way.  [Bolded emphasis mine.]

If the aircraft aren't approaching nearly or actually head-on, one aircraft will be unambiguously on the right:

As an extreme example, take the case of two aircraft "approaching" each other on side-by-side parallel courses in the same direction.  Obviously, in this situation, one aircraft will be on the right (and have the right of way).

(Red and blue are each a different scenario.)

Now consider two aircraft approaching each other on perpendicular courses (with a 90° angle between the two).  One aircraft's pilot will need to look to the left to see the other aircraft, and the other aircraft's pilot will need to look to the right to see the first aircraft.  The aircraft whose pilot has to look left is the aircraft on the right, and they have the right of way; the other aircraft is to the left of the first aircraft, and will have to give way.

Even if there's a 135° angle between the two aircraft's flightpaths, one pilot will still be seeing the other aircraft noticeably left of the center of their windscreen; this pilot's aircraft is on the right, and has the right of way.  The other pilot will have to look through the right side of their windscreen to see the first aircraft; the other aircraft is on the left, and has to give way.

I see, thanks so much! Very clear explanation - I realize I had mostly misunderstood the meaning of "converging" and that was the main cause of confusion

Good explanation. It also sounds like this rule was inherited from the "Stand on vessel" rule in navigation on water. In short : if the other vessel can see your green navigation light (also shared with aircraft) they have right of way.

Honestly, if the other aircraft is a B-2 Spirit Stealth Bomber, I'll probably yield right-of-way to them regardless of direction. And maybe take a picture.

I don't understand this sentence "But that's only the case if the two aircraft are approaching head-on, or nearly so, and the FAR you quoted already accounts for that situation:".  If the aircraft are nearly head-on but not exactly, it is still true that one (say "Aircraft A") will perceive Aircraft B to be slightly toward Aircraft A's left, while the other (Aircraft B) will perceive Aircraft A to be slightly toward Aircraft B's right.  As you in fact illustrated in your lower diagram.

In the US, a similar "guy on the right has right of way" rule applies to some traffic intersections, though I'm having a little trouble remembering which ones.  Not ones that are clearly marked as 4-way-stops, obviously, because then right-of-way is granted to whoever arrived at the intersection first.  I think the "guy on right has right of way" rule applies where two roads intersect with no stop signs.

@quietflyer ('s first comment): I'm assuming the FAR's mention of "head on, or nearly so" refers to cases when the two aircraft are so close to directly head-on that it isn't visually apparent whether the other aircraft is very slightly to the left or very slightly to the right.

@reirab: On the other hand, if your aircraft is an extraterrestrial flying saucer, they might want to yield right-of-way to you instead.

@quietflyer It is 4-way stops that you're thinking of, but it's only if the vehicles arrive at the intersection at the same time. If it's not obvious which one arrived first, the one on the right goes first.

@Vikki-formerlySean Hmm... Now this makes me wonder about this mid-air collision that occurred 20 years ago. 91.113 doesn't mention anything about priority of birds or baseballs and the bird was on the right, so I suppose it could be argued that Randy Johnson failed to yield the right-of-way and should have instead thrown a curve ball behind the bird's path. On the other hand, the dove is more maneuverable than a baseball, so it could be argued that the spirit of 91.113 would give right-of-way to Johnson's pitch.

I wonder if Randy Johnson got a number to call from FSDO about that incident? Or maybe just PETA?

Consider two aircraft converging at a 90° angle, as two cars arriving at a stop sign. "Rightness" or "Leftness" would be determined by the shortest angular distance between them: If B is 90° counterclockwise (right) of A, then A is 270° counterclockwise (right) of B—meaning A is 90° clockwise (left) of B. So B is "to the right" of A (as measured by the shortest angular distance) and B has the right-of-way.


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Is it possible to own an F-14 Tomcat?
Ok, I'm a wartime aviation enthusiast who loves, in particular, 3rd-5th generation fighters. I have held an avid interest in them for years, but I've always wondered whether or not I could own one myself. Obviously, it would cost a lot of money to own a fighter aircraft, and the aircraft would come with a lot of restrictions. But, my question is this: Is it possible to own the F-14 Tomcat (demilitarized/used) as an airliner/civilian pilot?
Although I am young right now, I plan to become an airliner pilot and eventually purchase a small plane (single engine turboprop) for myself. After all that, which I assume will take anywhere between 30-40 years, I would like to know if it would be possible to purchase and own an F-14. It is my all time favorite fighter for many reasons, the main one being its variable sweep wings. Assuming that I can acquire the proper training and equipment to utilize its true speed and maneuverability, would it be possible (in roughly 30-40 years) to purchase and fly an F-14 at its full potential, disregarding the costs to own and maintain it.

In the very unlikely event that a flyable F-14 will be available 40 years hence, it will be even more unlikely to get all spare parts to keep it flyable. But maybe there will be enough demand to create a modern copy - that happened before.

If you like variable sweep wings, a Tornado may be more practical.

Some were in private hands and it didn't go well:  https://www.latimes.com/archives/la-xpm-2007-mar-07-me-jets7-story.html

To its full potential?  Even if you had a working Tomcat, you couldn't fly it supersonic over or near land in most parts of the world, and I doubt the authorities in most places would look kindly on low-level high-speed terrain following exercises even in unpopulated areas.  Although IDK enough about aviation rules to be sure.   (IDK what you meant by "full potential".  I was assuming you weren't including the armament, and weren't planning to become an air pirate out of Porco Rosso.)

You may be interested in this video on Youtube: Badass Pilot Buys Own Fighter Jet. It's possible to buy military jets, but they don't let just anyone purchase or fly them. Additionally they're very hard to maintain.

I’ve chatted with a few F-14 pilots who are members of the Tomcat Association, aviation museum curators, as well as former NFWS Topgun instructor Dave “Bio” Baranak about the possibility of getting one or two F-14 airframes flying again.  The depressing conclusion we have all come to is that it is almost impossible for this to happen.  Multiple F-14 pilots, RIOs and maintainers have approached the DOD about doing so, to no avail.
Virtually all Tomcats which did not get donated to museums were cut up and melted down into beer cans, more or less.  The Navy destroyed nearly all of its Tomcat fleet to prevent spare parts from winding up in Iran, the last known Tomcat operator.  As of 2018, the DoD did maintain at least 11 flyable F-14 airframes in long term storage at AMARC and they are not parting with them.  Period.  The exact reason the DOD keeps these aircraft is unclear, though I suspect they are retained for limited purposes such as RedAir aggressor aircraft, flight test and research, or similar role that the type is deemed suitable for.  F-14s which were donated to museums or for public display as “gate guards” still remain the property of the DoD and the airframes have been mutilated to permanently prevent them from carrying flight loads.
The only other potential source of flyable Tomcats remains Iran and it’s believed that all or nearly all are out of commission due to lack of spare parts.  If somebody could purchase an F-14 airframe from Iran, then take it back to the states, it might be possible. However this would almost certainly be seized and confiscated by customs.
And even if you could get the aircraft back into the United States legally, plus find the enormous funding required for this enterprise, you’re still faced with the prospect of finding spare parts for it.  It will be very difficult to get anything from the manufacturer i.e. Northrop Grumman as the assembly lines have long since closed, there are ITAR and national security restrictions, and all existing spare parts, aside from those needed for the last AMARC F-14s, have been destroyed.
It’s a great dream, but it looks like if you ever want to see an F-14 fly again, the closest you will get is watching a DVD of Top Gun or Final Countdown.

The only other possibility might be a private company specializing in providing aggressor aircraft and pilots such as Draken. But I checked and they don't have F-14s in their inventory. Wouldn't make much sense anyway, their specialty is in aircraft the US doesn't have.

It’s possible, however very unlikely. For starters, the only source of airworthy Tomcats are the few AMARC birds left, which the DOD will not sell.  Second, it would be prohibitively expensive for a private firm to operate and far more economical options are available.

Would it be legal for a billionaire to start a company devoted to building new nonmilitary F-14s?  Or to pay Northrop Grumman to build one, assuming you could convince them?

So, when they make movies like Top Gun et al., does the DOD loan them out (along with qualified pilots) to the studios?  Or do they have to mock up fake ones?  Obviously now they can do it all with CG, but I suspect that was less viable an option when the first Top Gun came out.  (This might be more a question for Movies.SE)

"However this would almost certainly be seized and confiscated by customs." Yea, I wonder how many regulations you'd be breaking in such an attempt. Quite a few, I'd guess. As for seeing them fly, they're featured every so many episodes in JAG (the series) as well.

@Darrel Hoffman All military aircraft used in movies or other entertainment products are operated by military pilots and kept under the control of the DOD at all times. The military is perfectly happy to help Hollywood make movies because they’re free recruiting tools. The only exception is the DOD will not help out with projects that involve a character stealing military hardware as a plot point, which is why the producers of Iron Eagle had to get the aid of the Israeli Air Force for their F-16 and Kfir (MiG-23) footage.

@Ron Presser Theoretically you could approach any aircraft manufacture to build you an aircraft. The problem is that the F-14 engineering data package is ITAR restricted and requires congressional approval for the sale.  It’s also a proprietary trade secret of the Northrop Grumman Corp.  And it would be cost prohibitive to do so.

I believe the US Navy is particularly picky about ownership, too. They claim possession of even wrecks, which is part of why there have been no TBD wrecks recovered yet. The Air Force's policy is different, if it's before a certain time frame, and you can recover it, it's yours.

Also keep in mind that F-14s were retired largely because the Navy found then too expensive to maintain (yes, also because the threat they countered mostly ceased to exist). If the Navy can't afford them, it seems unlikely even an extremely well-funded private group could.

@Jerry Coffin Yes the airplane was expensive to operate and you would have to see if a case could be made for a business model of exhibiting a jet like that in airshows could pay for the maintenance and other operating costs.  I don’t think the labor would be difficult to find as I’m sure there would be a large line of former pilots, RIOs, and maintainers willing to donate their time and expertise to such a project.  If only the other obstacles listed here could be overcome.

@CarliFelicione Sustained funding is the problem. With a massive public fundraising effort, the Vulcan To The Skies group got a British Vulcan bomber flying again, and they did air shows for a couple of years. But as soon as the engines and airframe needed another overhaul, they had to retire it, because they couldn't do the same fundraising effort twice.

Highly unlikely.
There were only two operators of the F-14: the US and Iran. When the US decommissioned theirs in 2006, aside from a few stripped airframes in museums, they were all destroyed to prevent their parts making their way to Iran.
If Iran somehow has any flyable planes left in another 30-40 years despite not being able to get any parts to repair them (other than cannibalization) in the meantime, I doubt they would sell you one.


text:
Is it possible to continue flying without electric power?
To be specific with the question, I would like to know if if it is possible to continue flying in the event of total loss of electrical power. Obviously, I know this entails the shutting down of engines which would discontinue the ouput of thrust, therefor disabling the aircraft's ability to safely climb and maintain altitude. So, in technical terms, the aircraft would therefor be gliding rather than flying, but I would still like to know if its possible to do so without any power (primary or auxiliary/emergency, although having absolutely 0 power is extremely unlikely).
Here is my question: Is is possible to control an aircraft with absolutely zero electricity midflight? I also understand that smaller planes can be controlled purely by the hydraulics systems, but does that also apply to larger aicraft such as the Boeing 747, Airbus 320, etc.?

Answers vary widely depending on aircraft so you will need to be more specific.  Also, what leads you to believe that you would need to shut the engines down?

Gimli glider is a famous case of in-flight loss of electric power after fuel exhaustion, with a wind turbine used as a last resort generator.


text:
Why didn’t AFR4590 use heavy in its call sign?
https://aviation-safety.net/investigation/cvr/transcripts/cvr_af4590.php shows that AFR4590 didn’t use heavy in its call sign even though Concorde had a MTOW greater than 300,000 lbs.  Unlike BAW, AFR didn’t have a specific Concorde callsign, so why did they not say heavy:


text:
On IMA systems e.g. A380 how many people are required to preform regular maintenance checks on this system?
I am aware that there is usually 1 or 2 engineers preforming maintenance checks for a particular avionic system on an aircraft at a given moment. For IMA system, is there often more engineers working to get the maintenance checks completed. Are the maintenance check procedures/tests on IMA systems more complex that would require more people.

Not any more people than any other system. But your question raises a few key points for consideration.
First of all, maintainers don't actually go anywhere near the aircraft if it is not a requirement. I don't agree that there are usually 1 or 2 engineers on an aircraft at any given moment. If it aint broke, don't fix it. Over-maintenance is a cause of systems breaking more often due to factors such as human error, more cycles that could reduce life, etc. So unless there is a reason to be there, they won't be pushing buttons just to see if something is still working. In fact, your foreman/chief would probably ask you why you were touching that aircraft. Valid reasons to be working on the aircraft are most often that the maintenance is called for in the servicing schedule (scheduled maintenance), or it has broken (unscheduled maintenance). There are other exceptions to this including modifications and special technical instructions. These will have slightly different names depending on the local authority under which the aircraft operates.
Second point is that if you have been granted authority, have the current licence and the company policy allows for it, you may be able to perform maintenance by yourself. However, airliners usually (at least in Australia) have an AME (aircraft maintenance engineer) do the work and sign the first level of certification, followed by a LAME (Licenced Aircraft Maintenance Engineer) who certifies that the work was completed in accordance with the approved procedure and issues a certificate of release. This allows the aircraft to go flying again. In general aviation, it is more common for the LAME to do the work by themselves. In this case, they are kind of supervising themselves.
Third point: The complexity of the aircraft's system does not determine the number of "levels" of certification. If anything, it is how safety critical the system is to airworthiness that determines how many people must sign for the job. In some cases, an IMI (Independent Maintenance Inspection) or QA (Quality Assurance Inspection) is called for, increasing the levels to 3. Systems such as landing gear are pretty important so they may have 3 levels, but this isn't always the case. The levels are 1-AME, 2-LAME, 3-Independant. I don't think I've heard of Integrated Modular Avionics requiring 3 levels. As a training manager, I'd just make sure my people have adequate training and are confident. If they aren't confident, part of getting your LAME licence is knowing when to put your hand up and say you can't do something. Much safer to do that than guess what you are doing is correct. Hundreds of lives are at stake. You need to have the knowledge and maturity to be a good engineer.
Another interesting point worth bringing up: The Royal Australian Navy had a Sea King helicopter crash leading to loss of lives during Operation Sumatra Assist. This was for the 2004 tsunami. The cause was the lack of a split pin that should have been installed on the main rotor hub assembly. Long story short, the Navy increased the levels of inspection of this job (and others similar to it) from 3 to 4. Data showed that after this change was implemented, more mistakes happened, which is a bit counter-intuitive. You would think that by having 4 people look at something that it would be safer. Wrong. Data showed that people became more complacent because they thought that someone else would have their back if they were having a bad day. The Navy reverted back to 3 levels, and incidents reduced again.
IMA systems (I assume you mean Integrated Modular Avionics) aren't any more complex than aircraft without it. For anyone who is wondering what IMA is, it is basically a concept of merging what used to me a bunch of separate computer systems into a reduced number of boxes. That way the systems can share computing power. On the 787 these are called CPIOMs (Central Processing Input Output Modules). They've managed to reduce the physical footprint and weight of the entire avionics package than pre IMA concepts. There's actually a good argument to say that it is now simpler, because there are less boxes to maintain than before. System testing (BITE) is still done via an interface in the cockpit and is as easy as before.


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What is meant by the pointing vector and pointing vector magnitude in ECI?
In Siouris's Missile Guidance and Control Systems, it is stated that the angle of attack can be approximated in a 3DoF (degrees of freedom) simulation by the following equation:

I know that ECI refers to the Earth Centered Inertial Frame. The only thing I don't understand is what this pointing vector, and its magnitude meant to be. It is not the thrust vector because the units do not match up (m for meters, otherwise it would be N for Newtons which is used commonly in the book). And also, if it is a pointing vector, why would it even have units?

Could it be the direction in which the missile is pointing?

Indeed the pointing vector $R_{NT}$ is the direction that the missile is pointing in, as @Jim mentions in a comment. Which is not necessarily in the direction of the velocity vector, in which case the angle of attack would be zero.
The angle of the two vectors can be computed via the cosine rule, as explained in the wikihow. The pointing vector has units of length - it floats in space, but transplanted to the centre of the earth will determine a distance in 3 dimensions.
And the magnitude of vector (x,y,z) is $ \sqrt{x^2 + y^2 + z^2}$

Confirmed by the book: "The model assumes that the pointing vector and the missile body-axis are the same."

So does this mean that I can treat the rocket thrust vector as the missile pointing vector as a simplified assumption?

@alexmesa Yes indeed. On average, the thrust vector will point in the body axis.


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Is it possible for the lift force unit vector to be equal to the weight force unit vector?
I know that the drag force is parallel to the direction of flow, thus equal to the unit vector of velocity. But if you consider a vehicle, when its angle of attack is below the velocity vector from a local vertical local horizontal reference frame, does this mean that the lift vector will be pointing downward too and not up?

That would be a "spoiler".  Yes, of course negative AoA can produce downwards force.  Tails of many planes do the same (up or down) to control wing AoA.  Down force helps hold car on the road.

Is it possible for the lift force unit vector...

Be careful in the wording, the lift unit vector has by definition a magnitude equal to 1, and by convention positive lift is up (or in certain contexts oriented to the suction side of the airfoil, this is equivalent unless flying inverted).

when its angle of attack is below the velocity vector from a local vertical local horizontal reference frame, does this mean that the lift vector will be pointing downward too and not up?

In simple terms, your question boils down to: Can lift be negative? Yes lift can be negative by adjusting the angle of attack to a negative value.
However the angle of attack is not between the direction of flight (velocity vector) and the wing chord, but between the airflow direction (relative wind) and chord:

The two notions are only equivalent in certain conditions, like when flying at null or constant vertical speed.
Negative lift is is visible on the curve showing the lift coefficient:

Lift and drag coefficients according to angle of attack (α), source (adapted)
On the left hand side, the orange curve is below x axis, and is negative. This occurs for values of α < -3°. So for values >= -3° lift is positive, and below lift is negative, meaning a small negative angle of attack is not sufficient to orient lift downward.
There are control surfaces naturally developing a negative lift value: Horizontal stabilizers (on many aircraft). They are used to prevent the nose to pitch down due to the center of gravity being forward of the center of lift.

Horizontal stabilizers developing a negative lift, source (adapted)
In the figure above, lift W is positive. It acts on a point which is aft of the center of gravity CG (by dw distance), this creates a pitching moment, the nose tends to pitch down. Horizontal stabilizers are used to produce a negative lift T. Given the longer arm of T (dt), this small negative lift can counter the pitching moment due to lift W. When the aircraft is trimmed this way the wing generates lift W equal to the sum of weight and T.


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How can one change the wingtip geometry to enhance the wingtip vortices?
I'm doing some research on enhancing wingtip vortices as part of my Honours project. After spending a fair bit of time in the literature and on this forum I have some models which should kick-start my simulations. Because this is such a under-researched field (understandably) I wanted to mine the wisdom of those here: What design elements for a wingtip geometry would be conducive to creating a strong wingtip vortex?
The context is that I am studying how to enhance the efficiency of autonomous formation flight, using one drone as the 'tug-boat' for the other trailing drone. The leading drone would be outfitted with vortex-enhancing wingtips in order to increase the upwash experienced by the trailing drone. The overall purpose is to increase the range of the secondary drone at the expense of the first, in situations like scientific aircraft stationed in Hobart anf flying round-trip to Antarctica.
So far I have identified the Aspect Ratio, Lift Coefficient and Oswald Factor as parameters I'll need to change, based simply off the induced drag equation $C_{d_i} = \frac{C_L^2}{\pi*AR*e}$. This has given rise to potential solutions like using vortex generators on the wing surface to enhance lift while removing the winglet to allow the flow to curl around the tip, pitching the winglet up to increase its local angle of attack (effectively the same principle), and using a wingtip fence that runs from the leading edge to half the chord length and stops there, creating a high-pressure difference and then allowing the vortex to develop rapidly.
I'm aware I may have fundamental issues with this theory but also that I don't know what I don't know. If this won't work physically, then an explanation as to what I should be focusing on to create greater upwash outboard of the trailing edge of the leading wing would be much appreciated.

@Jpe61 Yes, his posts shaped my initial reading at the beginning of the year. And yep, those factors are ones I've seen many times now. With a lack of ability to change the wings, I'm essentially trying to find out if there's anything I can do (and if I should be doing it) with just the wingtip geometry.

You are not interested in wingtip vortices. For now you are barking up the wrong tree. For that tug idea to work, you need to focus on the trailing vortex which is a result of the whole wing, not the wingtip.
How to maximize trailing vortex intensity? Fly slow, use a small aspect ratio and create lots of lift. This means a heavy airplane with a high span loading, flying close to stall. Niceties like the spanwise lift distribution are much less important; all low aspect ratio wings will show an elliptical or nearly elliptical lift distribution anyways.
Next, position the following airplane right. Between the two vortex cores of the leading airplane you will find severe downwash. What goes up must go down. This is not only true for aircraft, but also for the air surrounding them. So any part of the following airplane needs to stay outside of the immediate wake. In order to maximize the recovery of the energy wasted in the leading airplane you need two following airplanes, one to the left and one to the right, and ideally both need to be specific to their side since both will fly in an asymmetric flow field. If you don't account for that by optimising wing twist and weight distribution, your rudder and ailerons on the following airplanes will do that, albeit less efficiently, just in order to stay inside the upwash part of the wake.
To answer what you asked: For maximum wingtip vortex intensity you need a steep gradient of lift at the tip. This means a spanwise lift distribution which peaks near the tip. Of course this is what a good airplane design will avoid at all cost, because it not only makes it less efficient, it also results in dangerous stall behavior.

So the optimal "tug" could be a lifting body no?

@JohnK Yes, but that one has a poor c$_{l:max}$. Maybe a delta wing or an F-104 would also do.

Many thanks, this is what I was beginning to suspect. To clarify: a steep lift gradient at the tip would create stronger wingtip vortices as a result of increased lift over the wing. This would do the same thing as "Fly slow, use a small aspect ratio" etc, but less well and more dangerously compared to altering the entire wing to ehance lift. Correct?

@HughesJC Not quite. The wing sheds vorticity over its whole span and the far-field vortex is independent of details like lift distribution, apart from the distance between the vortex cores. The near-field vorticity, however, can be increased locally by a steep gradient of lift over span, at the cost of less vorticity elsewhere. In all cases total lift will be the same. It is not about enhancing lift but to enhance the free vortex in the wake. And to help a following airplane you should focus on the far-field vortex.


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How much money can be saved by removing seats?
If have an airbus A380, and I remove 500 of the seats entirely from the airplane (Along with their overhead compartments), and the remaining 300 or so seats I fill up normally with people, is the MPG of Kerosene used by the plane substantially different?
Basically, assuming airlines could do it quickly (which right now they can't) would removing seats and overhead compartments from the plane be a substantial way to reduce costs on empty planes?
The limit of this question is also something I'm curious about. If I have an Airbus A380 completely stripped of seats and overhead compartments, how much better is its MPG compared to a normal one filled with people and baggage?

By the time you pay for the labor and certification costs you are way negative.  And you will probably never break even because you can't haul as many passengers, so what would be the point?

how much does certification cost? what is certification? Is it like an inspection that occurs before the plane is allowed to board passengers?

Depends a lot on the seat which had been installed. The newest economy seats weigh as little as 8 kg whereas a fully fledged business class seat can top 100 kg. For the rest please see this answer

Are you expecting to have 500 people standing and holding their carry-on luggage for the entire flight instead of sitting in their seats, or are you expecting to limit the flight to only 300 pax? 500 pax standing will never get certified (approved by regulatory agencies) because 500 loose projectiles would get hurt on take-off, landing, or during heavy turbulence.

@FreeMan no lmao we aren't soul plane, i'm just trying to think about what an airline could do if it has very few people booked on a flight. Those 300 ppl will get seats, the other 500 empty seats seem like such an enormous waste

Though removing the rear seat from my Piper Cherokee does make it a lot easier to carry my mountain bike :-)

@frogeyedpeas, I don't know what certification would cost, but it would be a one time deal to approve a specific configuration that you could then repeat.  (i.e. all cargo vs all passengers, or some hybrid in between...)  Think of it this way, do you remove the passenger seat from your car when you are driving by yourself to save gas?  And then reinstall it if you have a passenger?  If not, why not?  Because it would save some gas...

I don't know about the A380, but there are combi aircraft that have removable seats to use the remainder for cargo..

Sure, but that capability is part of their type certification, right?

The amount of time and manpower required to remove empty seats before a flight, transport them to the plane's next destination (presumably on another aircraft), then reinstall necessary seats for the next flight with 450 passengers (instead of this flight's 300 pax) would far exceed any potential cost savings of removing them for the flight. (Oh, don't forget, the remaining 350 seats would have to follow the plane to its 3rd destination because that flight is booked full.)
The biggest issue would be getting the seats themselves moved around so they could be reinstalled at the next destination. Not every aircraft uses the same seats, the mix of seats (cattle class, business class, first class) will be different on different versions of the same model of aircraft, even those owned by the same airline, so having a stock of seats at each airport that could be installed when the outgoing pax count is higher than the incoming pax count would be impractical.

It raises the intriguing and yet impractical idea of each passenger bringing their own seat.

Note that bush pilots in the likes of Cessna 208, Cessna 206 or Quest Kodiak often do add and remove seats as needed for the day. They don't add them for every flight, just in the install as many as the maximum passengers they are going to have on that round-trip, so the extra seats stay on the base. The free space can then be utilized for cargo.

The 208 seats 9, the 206 seats 6, and the Kodiak a whopping 10 @JanHudec. Removing all the seats, even working alone probably takes less than 30 minutes. Additionally, all 3 are probably making primarily round trips each day, and not flying Houston to Sydney. Also, carrying cargo in the space remaining doesn't lead to the fuel savings the OP was asking about. Not arguing that you're wrong, just pointing out the differences.

@FreeMan, indeed. I am not saying it's practical under the premises of the question—you are completely correct that it isn't. Just adding that situations, different from the question, where seats are added and removed as needed for the day do exist.


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Best way to forecast density altitude?
Let's say I'm planning a day trip from a low elevation home base to a warm, mountainous location. I wake up early, check the weather, do performance calculations based on current METARs, and decide to make the trip. However, when I go to return home in the late afternoon, I find the elevation and the heat make the density altitude too high for me to take off.
Using purely FAA-approved weather sources, how could this be avoided? As far as I can find, there are no forecasts that include temperature or pressure. Several non-aviation weather sites provide an hourly forecast of both of those things. Is that the best way to avoid being stuck on the ground at a high density altitude? It's obviously no substitute for a standard weather briefing, but it seems to be a necessary part of "all available information."

The best way is to check the weather and be very knowledgeable of the effects of weather on your altimeter.
Density altitude is primarily pressure altitude corrected for temperature, but higher humidity will also raise density altitude (H2O is lighter than N2 or O2).
The major factor is temperature, so taking off in early morning, when Temps are lowest, is plan 1.

Density Altitude = Pressure Altitude + [120 × (OAT - ISA)]

Also, keep an eye on lower pressure moving into the area, which will raise the pressure altitude around 1000 feet for every inch of pressure drop (one inch could be a fairly serious storm).
Humidity effects are minor, unless the air is very warm. At 10C air can carry 1% moisture, at 40C this rises to 5%, and at 50C to 9%!
But a big part of the "forecast" is how much weight you plan to be taking off with.  Leaving a few hundred pounds behind can greatly improve safety margains.


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Why doesn't Heathrow T5 have stand 504?
On the North side of Terminal 5A are four stands, 501, 502, 503, and 505. Why is there no 504?

And 510 is also missing…

Stand numbers at Heathrow correspond to the terminal and gate numbers. The first digit of the stand number is '5' for all terminal 5 stands, and the last 2 digits correspond to the gate number. So, the missing stands 504 and 510 would correspond to gates A4 and A10 respectively.
According to unverified (by me) reports on flyertalk.com, Gates A4 and A10 are exclusively "bus gates" where passengers board a bus rather than accessing via a jet bridge. So there is no corresponding stand next to the gate.

What is the purpose of a bus gate on A4 and A10, is it in case an arrival expected at the A gates gets parked remotely for some reason?

@opr Such "airside transfers" are used at many airports. As for the logistical rationale... I'm not really sure! I think that would make a great standalone question if you care to post it.


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What is a "runaway supercharger"?
During World War 2, my uncle was a navigator for a B-17G bomber based in the UK. On return from a bombing run over Frankfurt, the crew was forced to bail out over Belgium at 3000 feet altitude. My uncle and 3 of his crewmembers were rescued by Belgian resistance members and hidden for 8 months when Allied forces liberated western Belgium.
In the Escape and Evasion report the pilot gave on his return to 8th Army HQ, he wrote that the crew bailed out because "No 2 and 4 engines had runaway superchargers, lost oil. Would not feather with full oil pressure."
Could someone help me out here in understanding what's the likely causal sequence of events here? Why would the plane experience a "runaway supercharger?" Was that the cause of lost oil, or was it the other way around - some sort of oil loss causing loss of control of the supercharger?
Or was it that the propellers for 2 and 4 engines wouldn't feather properly (due to mechanical trouble or maybe flak damage?) causing the engine to overheat and burn oil?

Interesting question! Welcome to Av.SE.

and here I was envisioning a Tesla supercharger that'd grown legs... Bummer

The B-17 uses the term "Turbo Supercharger" which refers to just a Turbocharger (As John K points out in the comments, the former is actually the technically correct term). The B-17 had an issue with runaway Turbochargers they even cover it in the original training video (start @3:14). Its effectively when the waste gate gets stuck shut and you get excessive exhaust pressure driving the turbine. Along with the high pressure it causes the engine to lean out which presents even more issues.
The excessive oil burning could be from either the excessive pressure or the increased heat as you lean out due to turbo runaway.
The regulators also had manual control so you could deliberately shut the waste gate and create the same issue.

Turbo supercharger is the proper technical name for any turbocharger.

Dave, John, thank you both for these explanations and for the links to the videos. Super helpful and much appreciated. On the day my Uncle's plane went down, he was part of a group of 748 B-17s and B-24s, with fighter support from 637 P-38s, P-41s, and P-51s. Of the bombers, 23 planes were lost or were damaged beyond repair, and 359 were damaged. 7 fighter escorts were lost or damaged. Casualties were 7 KIA, 20 WIA and 203 MIA. Just on that one day.

Wow.  Yeah a mission loss rate of 3% isn't sustainable for very long. If you wanted to survive your tour in bombers, you wanted to be in a Martin B-26, with a mission loss rate only .5%.  Or a DeHavilland Mosquito which I believe was even less.  My father flew C-47s in Burma supporting the assault on Mandalay.

The B-17 had General Electric exhaust-gas-driven turbochargers (or more correctly, turbo-superchargers) that had pilot selectable boost levels. A selector dial in the cockpit controlled the operation of the wastegate that controls whether exhaust goes through the turbine and provide engine boost, or is bypassed directly outside.  The pilot's indication of the state of boost being provided to the engines would come from the engine manifold pressure gauges.
"Runaway superchargers" would suggest that the wastegate control systems on those engines were malfunctioning, such as the waste gate failing closed and directing all the exhaust flow through the turbo, causing the turbos to over boost the engines (the "running away" part), which would cause sky-high manifold pressures, detonation, and send the cylinder head temperatures through the roof.  This would necessitate a precautionary shutdown to keep them from blowing cylinders.
During the shutdown, even though they had oil pressure indication for the engines, the propellers, which operate hydraulically using engine oil, wouldn't feather. There is a feathering pump for each engine, normally electrically operated, that draws oil from somewhere in the system and can drive the prop into feather even without any engine oil pressure, as long as the pump itself has a source of oil to its suction inlet.
The turbosupercharger waste gate uses engine oil pressure for servo power for regulating the waste gate, that is also used to lubricate the turbo's bearings.
So... thinking it through, the turbo waste gate needs engine oil, and the feathering pump needs engine oil.  My guess is that there were oil system problems that impacted the waste gate control systems on those two engines, causing the boost runaway, and the same problem was also starving the engine's feathering pumps so that they were unable to complete the feathering of the propellers.  Maybe flak damage, or just a technical issue with the system (it would be odd for flak shrapnel to cause identical problems in two engines on opposite sides of the aircraft).
A B-17 at low altitude, well lightened, having burned off most of its fuel, dumped its bombs, and expended ammunition, would probably have been able to make it back to the UK on two engines, IF the two dead ones had feathered properly.  Having two unfeathered propellers windmilling away however, is massively draggy, and would have turned it into a two-engine glider, leaving the option of a forced landing or bailing out; and it looks like they chose the latter.

Possibly our old friend "same guy did maintenance on both engines, and did it wrong/used wrong parts", i.e. the reason they created ETOPS rules.  (Same guy can't do both engines).

The problem was far more likely to be enemy flak or fighters than a mistake by a mechanic.

Even that is pretty out of the box for that sort of double failure.  My guess would be a design or manufacturing defect what was relatively common.  The GE turbochargers were reputed to be somewhat sensitive (for example, the turbine could not tolerate any back pressure, which is why it was exposed).

This is speculative, but in diesel engines, if a turbocharger has a failed oil seal on the inlet side, the engine can then run on the oil leaking through the seal.
Since diesel engines are shut off by stopping the fuel supply, the engine then "runs away" and typically runs until it burns all it's oil, and seizes.
If these engines had a sufficiently high compression ratio, it's possible that they could auto-ignite on oil leaking into the intake.
This would support both the oil pressure issues and "runaway" description. I'm not sure if it's applicable to these engines in particular, though. Note that any engine can conceivably "diesel" if it has a sufficient compression ratio, or a fuel that ignites at a low enough temperature.
https://en.wikipedia.org/wiki/Diesel_engine_runaway

You mean they're not supposed to be diesel engines @ymb1 .  However, they are when they're dieseling...

I was ready to sell an anecdote about some oil that spilled into a cummins turbocharger during replacement. After it was cranked, the engine ran away- Turning the key off did nothing. Everyone ran for the hills. It ended up ejecting a connecting rod through the side of the block and out of the engine bay. Was hoping to learn about an airplane doing something similar.

@Harper-ReinstateMonica: Technically, when they're dieseling, they're Diesel engines (Diesel-cycle engines), but still not diesel engines (engines running on diesel fuel).

@ymb1 - The other answers covered a failure due to excessive intake boost pressure. I'm pointing out they could have failed due to unintentional dieseling and failed turbocharger seals. Any engine can be a diesel if you feed it a fuel that ignites at the chamber temperatures it achieves at maximum charge compression.

As with most aircraft, B-17 engines are low compression. I have never heard an aircraft engine dieseling.


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On average, how much fuel is left in commercial flights after landing?
I recently asked what kind of fuel reserve airlines were required to have. It turns out there are international guidelines for that, which for commercial flights are:

Per ICAO Annex 6, Part I, section 4.3.6 "Fuel Requirements," airplanes
should calculate their required fuel quantity as follows (summary; see
below for actual ICAO text):

Taxi fuel
Trip fuel (to reach intended destination)
Contingency fuel (higher of 5% of "trip fuel" or 5 minutes of holding flight)
Destination alternate fuel (to fly a missed and reach an alternate)
Final reserve fuel (45 minutes of holding flight for reciprocating engines, 30 minutes for jets)
Additional fuel (if needed to guarantee ability to reach an alternate with an engine failure or at lower altitude due to a pressurization loss)
Discretionary fuel (if the pilot in command wants it)

Considering a flight has many unknown, the same connexion using the same airplane will have different fuel requirement and fuel remaining, simply depending on the date you decide to have a look at the data.
Is there any statistics of how much fuel is left on average in aircrafts after landing of commercial flights?
It can be an average for commercial flights in general, a breakdown of short / medium / long haul flights, expressed in percentage of tank volume or in liters / gallons / weight. Or any usable measure I did not think of that can be used to do basic calculations.

There are way too many variables here. For example some regional aircraft may carry extra fuel to get to a destination that has cheaper prices or a contract. I also don't know that an accounting is done in that regard, usually the pilot just asks for a specific amount for the next leg (or more).

If fuel costs more at airport B than at airport A, and the cost of carrying extra fuel from airport A to airport B is less than the difference between the cost of the fuel at A and B, then airlines will sometimes carry more than they need to save money overall.

@RonBeyer I agree there are many variables, I'd be interested to know if statistics exist. We could get an estimate using fuel economy figures like the one in this page: en.wikipedia.org/wiki/Fuel_economy_in_aircraft ; but that wouldn't take into account other externalities like cheaper fuel you mentioned.

Some airlines fill the fuel tanks to the brim if it's going to make many short hops, the faster turnaround offsets the extra fuel burn and wear on the aircraft. I believe this would mess with the data you'd have with averaging the fuel left in arriving aircraft.

@EfeBallı Good to know. Cost of aircraft remains low compared to fuel. I guess this practice is more common when the crew of the plane remains the same (when labor is the highest operating cost, which it is for some high salary countries). Operating costs source (usa 2019): https://i2.wp.com/transportgeography.org/wp-content/uploads/airline_operating_costs_usa.png?w=1000&ssl=1

Also, you can do more flights and carry more passengers in a given time period.

The short answer is that airlines aren't required to report fuel on landing so there is no central database of average landing fuel.

P.S. and even if there were, it would be kinda meaningless due to differences in size aircraft, distance to diverts, possible tankering for the next leg, etc.

For a 737, between 5,000 and 8,000 lbs of fuel left at landing is fairly typical. For a short flight in good weather to an airport with multiple runways and nearby divert fields, the lower value is more likely; for a longer flight, and/or bad weather where holding is more likely, and the need to have fuel available for a possible divert, the higher value (or more, sometimes) becomes more likely.
That leaves aside the possibility of fuel tankering, which is adding more gas than the flight needs, so that less (or no) fuel has to be bought at the destination airport.  If fuel is expensive at the destination (compared to the departure airport) and the passenger/freight load is light enough, it's possible to land with enough fuel in the tanks that no refueling is needed to fly the next flight (assuming it isn't too long).  Tankering for differing fuel prices is referred to as "economic" fuel tankering; there is also "operational" fuel tankering, which occurs when fuel at the destination is either unavailable (natural disaster, perhaps) or running low (delayed delivery, some tanks/pumps inop, etc).

The usual reserve is 10%. On QF9 Perth to London we depart with 110,000 litres and land with between 8,000 to 10,000 litres remaining.

While relevant, one data point is not really a statistic. Do you have any additional information?


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For commercial flights, is there regulations imposing a minimum fuel reserve upon landing?
In aviation, weight is tightly linked to efficiency. The more fuel you take, the more weight you add and the less money you earn. On the other hand, you want to pack more to deal with unexpected headwinds or busy airports.
Following the 2008 economic crisis, banks were legally forced to increase their cash reserve to be able to withstand more uncertainty.
I understand that you get a priority landing spot if you're low on fuel but it only works when it's a marginal event. With fuel being one of the major cost of airlines, the pressure put by low cost operators, a will to reduce CO2 emissions, the advent of statistical analysis, it would be logical to see this reserve shrink with time until an accident happens; unless there is a legal limit no operator would dare going under?
I have no clue how well internationally-coordinated airlines regulation are (I would say not so good as the 737-max re-certification is done separately at least in the US, Europe and China), so any country specific value or international guideline is appreciated.


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how is ITT peak temperature affected by altitude during an airstart?
another T-6 Texan II (turboprop aircraft) question. When reading the airstart prodedure in the engine failure during flight section of the manual, I found something that Id appreciate the clarification. It mentions, "The higher the altitude, and the slower the airspeed, the warmer the starting ITT peak temperature."
By airspeed, I believe it is referring to KIAS (all graphs and useful data for pilots is in KIAS, at least in this manual) so no arguments there. now, we know that the higher the altitude the less dense air gets to keep up with a relatively constant KIAS, then TAS has to increase.
If TAS increased at this higher altitude, (relative to the air the plane is passing faster) and the plane is making up for the less dense air, why would we expect a higher peak ITT compared to an airstart at a lower altitude? Shoudnt be about the same peak temp?
Also the higher you go the colder it gets at least for a good 20,000 ft MSL which is a major portion of the flight envelope for this aircraft. (max ceiling 31,000 ft MSL). I guess getting colder air has zero impact on ITT? I appreciate the discussion. TY all!

Could you tell if RPMs are higher.  It seems the compressor needs to work a little harder with lower density and airspeed.  Agree with answers, but the InterTurbineTemperature may be higher if more fuel needs to be burned to run the engine.  Just curious about this one.

ITT is affected by the heat dissipating effect of airflow:  The less air flowing through the turbine section the hotter it can get.
So, you are correct in that IAS is the parameter that is being referred to by term "airspeed".  However, I think you are confusing things by bringing TAS into the discussion.  Keep the same unit of measure - if IAS increases then there will be a higher volume of air, and more heat dissipation; lower IAS means less airflow and higher ITT.
The reason you might get higher ITT at the same IAS but a higher altitude isn't because of TAS, it is because the air is less dense.  (yes, I understand that TAS and density are related...)  Air has thermal mass, and the less dense it is the less mass there is available to absorb heat from the turbine.
It might seem odd as cold as it is up high, but when performing environmental qualification testing of electronic equipment we often had more difficulty with heat dissipation during the high altitude chamber tests, (despite the sub-freezing temperatures) simply because the air was so thin.

Heat dissipation is roughly proportional to mass flow times temperature difference.
Mass flow through the intercooler is $\dot{m}=\textrm{TAS}\times \rho$. Since $\textrm{TAS}\approx\textrm{EAS}\sqrt{\dfrac{\rho_0}{\rho}}$ we get that $\dot{m}\approx\textrm{EAS}\sqrt{\rho_0 \rho}$. In words: the effect of increasing TAS and decreasing density don't exactly cancel out.
The temperature at altitude is only a few tens of degrees lower whereas the turbocharged air is a few hundred degrees. So not much of a factor.


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Why are spins not allowed below 10k ft msl in the T6b?
Reading through the manual I see that spins are not allowed below 10000 msl in this turboprop aircraft, the t6b, the reason that is written in the manual is as follows, quote:
"Spins below 10k msl are prohibited due to high stress on the propeller which occur during the spin maneuver with the propeller RPM below 80%. To prevent high stresses from occurring, the power management unit (PMU) artificially maintains propeller RPM at 80% with the PCL at IDLE when the aircraft is above 10000 ft pressure altitude."
A bit of background on the maneuver, it is executed with PCL idle airspeed 150kts and 30 degrees nose up. So I understand that at low altitudes the air is much denser (below 10k) and as the aircraft comes down with PCL at idle apparently at or above or less? (I dont know) 80% rpm the air would push the blades more than the rpm that is giving out by the engine? Like turning more than is supposed to, damaging the engine (its designed to move the air and not the prop being moved by the air)? I am missing I guess a/some concepts with torque RPM and propeller RPM? If someone please would help me understand why or how below 10k could damage the engine/prop during a spin I would be very grateful. the engine used is a Pratt & Whitney Canada PT6A-68 turboprop, 1,100 shp (820 kW). there is only one lever to control the power (PCL), it is a variable pitch propeller but its automatic, we only control the PCL for torque power the PMU and the Propeller interface unit (PIU) control the speed (Np) and blade angle

Just an editorial comment:  The quote in the second paragraph answers the question.  Perhaps consider rephrasing to something like "what causes propeller overstress in a spin below 10K'?"  The rest is good background as to why you are asking...

Here's my take.
The propeller is under huge centripetal forces when at operating speeds, and this force is essential for keeping the blades from bending too far forward (thrust) or against the direction of rotation (drag). This is why most helicopters have some sort of a hinge mechanism, they let the blades free and allow them to position themselves in a cone for the thrust and centripetal force to balance each other out.
Reducing prop RPM and placing the aircraft in a spin does 2 things.
-It reduces centripetal force which is necessary for placing the force vector on the prop blade axis
-Airspeed drops and angle of incidence between the prop disk and incoming airflow is varied wildly. This increases thrust (props generate the most thrust at zero airspeed) and places loads on the blades and thus the propeller shaft under loads that vary in magnitude and direction.
These loads can easily propagate through the reduction gear and the free turbine shaft.

Thank you for the comment, do you think you could draw the angle of incidence you are talking about in relation to the prop disk and incoming ariflow. I am trying to picture what you said in my mind. Thank you for your time Efe!


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How does the weight affect the altitude that can be reached by trading speed after a total engine failure?
I recently read in the manual for The Beechcraft T-6 Texan II (a turboprop aircraft) about engine failure during flight at different altitudes and the gains from pitching up after the engine failure occurs to trade airspeed for altitude.
They show graphs that depict the altitude gain from pitching 20° nose up attitude bleeding airspeed from 200 KIAS to 145 KIAS at different Pressure Altitudes. The not so intuitive part is that shows how the heavier the aircraft gets (by having more fuel) the more altitude gain by pitching the same amount of nose up (20°) until airspeed bleeds off from 200 to 145 KIAS in case of engine failure. could someone please explain how could this be possible? I doubt there is a mistake in their manuals. There are a couple of these graphs from 250 to 145 and 200 to 145 at different Pressure altitudes.
At a first glance, it may seem intuitive to think that, if the heavier aircraft encounters an engine failure and it pitches up (same amount nose up attitude as the lighter aircraft) it will produce the same amount of lift as the lighter one (at the same pressure altitude and same airspeed) but with more weight the less lift to bring the aircraft to a higher altitude compared to a lighter aircraft. I guess that is not right?
If there is a vector analysis/equations or any math proving this I would appreciate it a lot.
In other words: two identical turboprop aircraft, except A weighs less than B, fly at same KIAS, engine fails, both pitch up 20° nose up, which one gains more Alt?
The graph of weight vs altitude gain:

To gain altitude (by any means) you need to gain potential energy. Since the engine has stopped, the only thing you can do is trade kinetic energy for potential energy (slow down).
Since both kinetic energy ($E_k = \frac12 m v^2$) and potential energy ($E_p = m h g$) are proportional to mass, for a spherical cow trading the same speed would result in the same gain of altitude independent of its mass.
But an airplane isn't a spherical cow. There is one more force we need to account for: the drag. Drag has two components:

Induced drag (the cost of producing lift) is proportional to lift which is to sufficient approximation equal to weight (at 20° the approximation that $\sin x = x$ is still good enough).
Parasitic drag (the cost of moving through the air), however, is not proportional to lift; it depends only on the speed, air density and shape of the aircraft.

Since all the variables except parasitic drag are proportional to weight, the parasitic drag will take relatively less energy for a heavier aircraft (more energy to begin with), indeed allowing the heavier one to climb higher.

For a similar case that might be a bit more intuitive: consider having two balls of similar size, but significantly different weight. A ping-pong ball and a golf ball may be a good example, because they are light enough that accelerating your hand becomes the limiting factor, so you can throw the lighter ping-pong ball as fast as the golf one, but not much faster. But the golf one will then fly far, while the ping-pong one will slow down and drop much closer. This is because the air drag is relatively much bigger for the lighter ball.

Your ping pong and golf ball comparison really nailed it for me!

@Jan Hudec great thoughts TY. When you mentioned that "Induced drag (the cost of producing lift) is proportional to lift which is to sufficient approximation equal to weight..." I do see how induced drag is L*sinx and 20 being small (pi/9 ~ 0.3) might be a small amount to consider but what does weight has to do with them. are you saying the heavier one will create proportionally more lift? not sure. I thought both examples produce the same lift because of the same pitch attitude.

@Jan Hudec What about conservation of momentum? do you think that is why the parasite drag affects the lighter one more than the heavier one? therefore taking energy from the system at a less rate or less compared to the lighter aircraft.
Lastly, extra bonus follow-up question. once both aircraft reach their max altitude gained, do you think the heavier one will have a greater descent rate?

No. The reason parasite drag affects the lighter aircraft relatively more is that drag is taking away energy – drag is force and force times path is work, that is energy – at the same rate, but there is less energy to begin with. Conservation of momentum is a reason why induced drag exists, and why it is proportional to the lift (and decreases with wing span).

@YamchaAviator It's basically just that a heavier aircraft has more energy for the same speed than a lighter aircraft (the heavy one will have had to have already burned more fuel to gain that energy/speed). So an effect that reduces the aircraft's energy by a magnitude that does not depend upon weight will take less speed from the heavier aircraft than from the lighter one.

At least to me, it looks like the chart is missing 1000lb of weight values on the x-axis between 6900 and 8100 lb. Doesn't that imply that there should be a discontinuity in the altitude gain at that point? Instead it's drawn as a linear progression, which makes no sense. That doesn't help someone grasp this!

A few issues with the concept unless hard test flight data is shown.

Kinetic Energy vs drag

Glider data shows us that two gliders of different weight will glide the same distance.  The heavier one must glide faster but has more potential energy.  The two perfectly cancel each other out.
So Vbg for the fuel laden craft will be higher, therefor it must break off its ascent sooner.

both craft must turn to ascend

Obviously, the heavier craft must either pitch harder or fly a larger radius to reach 20 degrees nose up
Once at 20 degrees nose up, both parasite and induced drag will be greater because the entire plane has a higher AoA.
Procedurally, yes, pitch up for Vbg, but be aware of what it is for your weight.  145 knots may be too slow for a plane with a large fuel load.

Do you think conservation of moment has to do with the energy that is stored in the system at the beginning of the climb? that is why drag has a larger effect on the lighter aircraft (less mass, less momentum) ..maybe?

@YamchaAviator Yes, Jan Hudec answer is correct.  I wouldn't go down to 145 if heavy.  The real key to Vbg is optimum AoA.  The heavier plane should break at a faster airspeed.  Really wanted to see curves.  They are linear.  Thanks!

TY for the input! just as the last thought once they have reached max altitude and start descending, which one do you think will have a greater descent rate ?

@YamchaAviator theoretically, if both planes can hold optimal AOA, both planes descend at the same angle (going just as far) but (by geometry) the heavier plane has the faster descent rate.  But, for Vbg, it's the angle that counts.

@RobertDiGiovanni so basically wrapping it up as a lesson learned today. there is no perfect world, the lighter aircraft might get a slower descend rate while not as high as a climb than the heavier one but the heavier one will have a higher descend rate. would you agree? thank u!

@YamchaAviator yes, and thanks for an excellent question.


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is pressure altitude at different temperatures (different days) on the ground at the same airfield the same?

let's say we are at x-airfield which is at 0 MSL on the ground. We set the Kollsman window to 29.92 on a standard day, whatever the read is that would be the Pressure altitude, correct? (ideally zero too?). The next day is a very hot day and I set it to 29.92 would that give the same indication from the previous day?

if the pressure altitude changed from the previous day, I would say it was because of temperature, correct? however, when reading about it, it says that:
"Because Pressure altitude is a measure of the weight of the air above you. Temperature does not affect that. Heating the air just makes it expand"
it seems contradictory to me, saying PA wont change due to temp but if in the airfield the temp is different, it would read a diff PA, right?
I would appreciate the explanation. Thank you

First of all, the pressure at sea level varies from day to day. Pressure systems are moving over the earth, driven by interactions of temperature differentials in the atmosphere, temperature differences between the atmosphere and water within oceans and lakes, the influence of upper-level disturbances, as well as the amount of solar heating or radiationized cooling an area receives.
The ICAO Standard Atmosphere (ISA) assumes a sea level pressure of 1013.25 hPa (equal to 29.92 inches of Mercury or inHg). ISA also assumes a temperature of 15°C at sea level, with a lapse rate of -6.5° C/ 1000 m of geopotential height. This is what the altimeter is calibrated to.
If you are at sea level on a ISA standard day and altimeter is set to 29.92 inHg (1013.25 hPa) then the altimeter will read 0 ft.
If the temperature increases, but the sea level pressure remains standard (1013.25 hPa) then the altimeter will still indicate 0ft.
Only if the sea level pressure is not standard, then an altimeter (set to the standard pressure of 29.92 inHg/ 1013.25 hPa) will not read 0 ft at sea level. For every hPa of deviation, the altitude will change by approximately 27 ft.
Now it gets more interesting when the airfield is on top of 1000ft cliff next to the sea.
On the ISA standard day, the airport will report a QNH of 29.92 inHg / 1013.25 hPa. The altimeter set to that value will read 1000 ft at the airport.
On the hot day, with sea level pressure at 1013.25 hPa, but a sea level temperature of 35° Celsius (ISA +20° C), the same altimeter setting would result in an altimeter reading of approx 935 ft. This because the high temperature causes the atmosphere to expand, and the pressure at which 1000ft would be found in the standard atmosphere, is now elevated to approximately 1065 ft.
The meteorological station at the airport, will assume that the atmosphere is standard. Based on the actual pressure measured at 1000ft, they will calculate what the sea level pressure is, based on the assumption that the temperature is standard. And thus the QNH will be reported as 29.99 inHg / 1015.65 hPa, and at this setting the altimeter will read 1000 ft again.
When we now land at the sea with this setting of 1015.65, the altimeter would read approximately 65 ft, because in reality the pressure at sea level was 1013.25.

thank you for your answer. it seems like I go back to my question, why is it that pressure can stay constant at sea level (1013.25) but temperature can change? ... so are you saying pressure is always 1013.25 at sea level?

also in your last example, how is it possible to have 1013.25 at sea level and 1015.65 at 1000 ft?
when you say "..now land at the sea with this setting, the altimeter would read 65 ft approx" which setting are you referring to 29.99 or 29.92?

@YamchaAviator, the answer starts by clearly stating that pressure varies due to weather (due to wind patterns), even at sea level. And its lapse rate also varies with weather (with temperature). In the last example, the actual pressure at 1000 ft is 979.5 hPa, not 1015.6 hPa. The 1015.6 hPa is the actual pressure, plus lapse rate over 1000 ft the altimeter is supposed to show. But the altimeter does not have a temperature setting, so it uses the higher lapse rate for standard temperature. Since it's warmer, the lapse rate is lower. That's the point of the image you quoted in the question.

"Because Pressure altitude is a measure of the weight of the air above you. Temperature does not affect that. Heating the air just makes it expand"
The above statements are true but extremely misleading in my opinion. I cringe every time I see pressure altitude described like that and I understand why one might get confused.
Pressure altitude is just a way to express a pressure as an altitude. When you see a pressure altitude think of it as a pressure because that's all it is.
Imagine that instead of saying that your height is 6ft, you say that your "height frequency" is 0.37 Hz, because that's the frequency a simple 6ft pendulum would have. It would just be an equivalent way to express your height. That's what pressure altitude is.
So in a way it's true that temperature does not affect pressure altitude because you don't factor in temperature to compute pressure altitude. But temperature does affect pressure so you can see why it's confusing.
In the same analogy it's like saying that age doesn't affect height frequency because you just need your height to compute your height frequency... But your height frequency was not the same 40 years ago (assuming you're older than 40yo) so in a way, age does affect it.
So to answer your question you could get the same reading on a cold day and a hot day but very likely you won't. It's like asking "is the pressure the same on a cold day and on a hot day?" It could be... But most likely it isn't.

I think the confusion lies in the altimeter setting or kollsman window.
The airport altimeter setting is designed to ensure the airplane's altimeter will read field elevation (within 75 feet).
That airport altimeter setting is affected by both pressure and temperature.
The aircraft altimeter is designed to mimic the lapse rate of a standard day.  Roughly 1" mmHg and 2° c per thousand feet.
This is the key point.  If the aircraft altimeter is at a higher elevation than where the airport altimeter setting was read, the altimeter will show an error as in the chart you posted.  If the aircraft is at the airport field elevation, temperature has no effect on the altimeter as temperature has already been calculated in the airport altimeter setting.
This might make you think that temperature effects are increased with deviation from the airport elevation... and I would agree.  The FAA has a discussion of cold temperatures on approaches and the effects it has on the altimeter.  There is a correction chart for temperature on the altimeter.  The correction is based on the temperature and the altitude above the airport altimeter setting.
AIM Chapter 7 Section 3.
https://www.faa.gov/air_traffic/publications/atpubs/aim_html/chap7_section_3.html
You can see the effects of temperature on the altimeter setting by finding close geographic weather stations at quite different elevations.  It is quite pronounced by examining the Colorado airports and comparing them to the weather stations on top of the mountains.  There is a very short distance between the stations but quite different altimeter settings.


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where is the centrifugal force the flyweights experience coming from?
The T6b uses a PT6-68 engine which has a variable pitch, constant speed, feather, and non-reversible prop system. There is only one lever in the cockpit which regulates Torque and the Power Management Unit (PMU) whenever is on (PMU switch on, under normal operations) would regulate the pitch of the propeller in order to keep the propeller at a constant speed 2000 RPM (the Reduc Gear Box lowers it from about 30k RPM to 2k)
Here is the manual for the P&W PT6 series I found online:
https://mikeklochcfi.files.wordpress.com/2018/08/training-pt6a-60-series.pdf or you can google "pt6 60 series manual"
Under the Propeller system (page 216 pdf), I believe the PMU On mode is referred to as the "governor governing mode".

"When the engine output power is increased, the power turbines tend to speed up. The flyweights in the CSU sense this acceleration. The flyweights go into an overspeed condition because of the increase centrifugal force and force the control valve to move up and restrict oil flow to the propeller dome. The feathering spring increases the propeller pitch (more towards feather position) to maintain the selected speed"

First, What makes the Flyweights spin up? There is no blue lever in this plane. I read from other designs that flyweights are on top of a disc sometimes, what makes it spin though? I don't see how the engine spooling up is connected to the flyweights. The counterweights on the other hand are spinning because are connected to the spinning hub, but the Governor in which the flyweights are is not spinning I think. In the picture, the flyweights are the yellow u-shape bars above the Pilot Valve
I also have a second question regarding this propeller system I posted separatelly. Thank you

I see where your confusion is...  Yes, the CSU does NOT spin, but there is a 90 degree gear and vertical shaft running into the CSU that spins these flyweights.  I don't have time for a full answer right now, but look at the diagram on 11.5 and you see a crude representation of these gears.

Sir, I found it thank you. Yes I see now there are a couple of gears that are connected coming from the main turbine shaft. I posted another question about this topic, you could find it in my profile Im sure. Id def appreciate the help please